Take, $c$ represents a constant and $x$ represents a variable. The derivative of constant $c$ with respect to $x$ is written in the following mathematical form in differential calculus.

$\dfrac{d}{dx}{\, (c)}$

According to definition of the derivative, the differentiation of $f{(x)}$ with respect to $x$ can be written in limit operation form.

$\dfrac{d}{dx}{\, f{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{f{(x+\Delta x)}-f{(x)}}{\Delta x}}$

Take $f{(x)} \,=\, c$, then $f{(x+\Delta x)} \,=\, c$. Now, substitute them in this formula.

$\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{c-c}{\Delta x}}$

Now, take $\Delta x = h$ and express the equation in terms of $h$ from $\Delta x$.

$\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{c-c}{h}}$

There are two terms in the numerator and they both are equal. So, the subtraction of them is equal to zero.

$\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{\cancel{c}-\cancel{c}}{h}}$

$\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \Big(\dfrac{0}{h}\Big)}$

The quotient of zero by $h$ is zero mathematically.

$\implies$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize (0)}$

Now, calculate the limit of zero as $h$ approaches zero. In this case, there is no $h$ term in the function and the value of the function is zero. Therefore, the differentiation of a constant with respect to a variable is equal to zero.

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, (c)}$ $\,=\,$ $0$

In this way, the derivative of a constant rule is derived by first principle in the differential calculus.

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